How do you integrate #int cosxsqrtsinx# using substitution?
1 Answer
Dec 18, 2016
Let
#=int(cosxsqrt(u))/cosx du#
#=sqrt(u) du#
#= u^(1/2) du#
#=2/3u^(3/2) + C#
#=2/3(sinx)^(3/2) + C#
#=2/3sqrt(sin^3x) + C#
#=2/3sinxsqrt(sinx) + C#
Hopefully this helps!
